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प्रश्न
Find the sum of first 16 terms of the A.P. whose nth term is given by an = 5n – 3.
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उत्तर
Given, nth term of A.P. is an = 5n – 3
∴ a1 = 5(1) – 3 = 2
And a2 = 5(2) – 3 = 7
∴ Common difference, d = 7 – 2 = 5
∴ Sum of n terms of A.P., Sn = `n/2[2a + (n - 1)d]`
∴ S16 = `16/2 [2(2) + (16 - 1)5]`
= 8(4 + 75)
= 8 × 79
= 632
संबंधित प्रश्न
The sum of three terms of an A.P. is 21 and the product of the first and the third terms exceed the second term by 6, find three terms.
Find the sum of the following arithmetic progressions:
`(x - y)/(x + y),(3x - 2y)/(x + y), (5x - 3y)/(x + y)`, .....to n terms
Find the sum of all integers between 100 and 550, which are divisible by 9.
How many terms of the AP 63, 60, 57, 54, ….. must be taken so that their sum is 693? Explain the double answer.
In an A.P. the 10th term is 46 sum of the 5th and 7th term is 52. Find the A.P.
The sum of first 9 terms of an A.P. is 162. The ratio of its 6th term to its 13th term is 1 : 2. Find the first and 15th term of the A.P.
If the sum of P terms of an A.P. is q and the sum of q terms is p, then the sum of p + q terms will be
If the sums of n terms of two arithmetic progressions are in the ratio \[\frac{3n + 5}{5n - 7}\] , then their nth terms are in the ratio
How many terms of the A.P. 25, 22, 19, … are needed to give the sum 116 ? Also find the last term.
Find the sum of natural numbers between 1 to 140, which are divisible by 4.
Activity: Natural numbers between 1 to 140 divisible by 4 are, 4, 8, 12, 16,......, 136
Here d = 4, therefore this sequence is an A.P.
a = 4, d = 4, tn = 136, Sn = ?
tn = a + (n – 1)d
`square` = 4 + (n – 1) × 4
`square` = (n – 1) × 4
n = `square`
Now,
Sn = `"n"/2["a" + "t"_"n"]`
Sn = 17 × `square`
Sn = `square`
Therefore, the sum of natural numbers between 1 to 140, which are divisible by 4 is `square`.
